The radius of a circle is r yards. Is the area of the circle at least

Firstly notice what the question stem is asking for. Is πr^2 ≥ r? [ Note: π=pi]
We can simplify it more. πr ≥ 1
=> Is r ≥ 1/π? The question is stated in yards, but the statements use feet, so we'll have to convert them.
Given: 1 Yard= 2 Feet
Thus, r ≥ 3/π (now it's in feet)

The value of 3/π is approximately\(\frac{3}{3.14}\) which is around 0.955, i.e., a little less than 1, so we'll consider it as 1.

So ultimately, we need to find whether r≥1?

(1) Given: Diameter>2 feet => 2r>2 =>r>1 SUFFICIENT

(2) Given: r=2ft and πr^2 > 2f
Simplify πf^2 > 2f
f > 2/π
f > 2/3.14 (~0.65)

f could be smaller than 3/π feet, i.e, 1 but it could also be larger. Nothing about f is mentioned in the question if you notice. INSUFFICIENT


I hope you'll get it now.

This is MUCH harder than I expected. Excellent question (kudos)!

\(\pi {r^2}\,\,\mathop \ge \limits^? \,\,\,r\,\,\,\,\left[ {{\rm{yard}}{{\rm{s}}^{\rm{2}}}} \right]\,\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{r\,\, > \,\,0} \,\,\,\,\,\,\,\pi r\,\,\,\mathop \ge \limits^? \,\,\,1\,\,\,\,\,\left[ {{\rm{yards}}} \right]\)


\(\left( 1 \right)\,\,\,2r\,\,{\rm{yards}}\,\,\, > \,\,\,2\,\,{\rm{ft}}\,\,\left( {{{\,1\,\,{\rm{yard}}\,} \over {3\,\,{\rm{ft}}}}} \right)\,\,\,\,\,\,\left[ {{\rm{yards}}} \right]\,\,\,\,\,\, \Leftrightarrow \,\,\,\,r > \,\,{1 \over 3}\,\,\,\,\left[ {{\rm{yards}}} \right]\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{ \cdot \,\,\pi } \,\,\,\,\,\,\pi r > 1\,\,\,\,\,\left[ {{\rm{yards}}} \right]\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)


\(\left( 2 \right)\,\,\,\pi {f^2}\,\, > \,\,2f\,\,\,\,\left[ {{\rm{fee}}{{\rm{t}}^{\rm{2}}}} \right]\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{f\,\, > \,\,0} \,\,\,\,\,\pi f > 2\,\,\,\,\,\,\left[ {{\rm{feet}}} \right]\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\pi f\,\,{\rm{ft}}\,\,\,\left( {{{\,1\,\,{\rm{yard}}\,} \over {3\,\,{\rm{ft}}}}} \right) > \,\,\,2\,\,\,\left( {{{\,1\,\,{\rm{yard}}\,} \over {3\,\,{\rm{ft}}}}} \right)\,\,\,\,\,\,\left[ {{\rm{yard}}} \right]\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,{{\pi f} \over 3} > {2 \over 3}\,\,\,\,\left[ {{\rm{yard}}} \right]\)

\({{\pi f} \over 3} > {2 \over 3}\,\,\,\,\left[ {{\rm{yard}}} \right]\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{r\,\, = \,\,{f \over 3}\,\,!!} \,\,\,\,\,\pi r > {2 \over 3}\,\,\,\,\left[ {{\rm{yard}}} \right]\,\,\,\,\,\,\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\pi r = {3 \over 4}\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,\pi r = 1\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Statement 1:
Test the threshold value: d = 2 feet, implying that r = 1 foot = \(\frac{1}{3}\) yard
Case 1: \(r = \frac{1}{3}\) yard
In this case, area \(= πr^2 = π(\frac{1}{3})^2 = \frac{π}{9} =\) more than \(\frac{1}{3} \)square yard
Since the area is greater than r square yards. the answer to the question stem is YES.

Test a greater value: d = 6 feet, implying that r = 3 feet = 1 yard
Case 2: r = 1 yard
In this case, area \(= πr^2 = π1^2 = π =\) more than 3 square yards
Since the area is greater than r square yards. the answer to the question stem is YES.

Since the answer is YES whether r is at or above the threshold, SUFFICIENT.

Statement 2:
Since the circle has a radius of f feet, the area of the circle is \(πf^2\) square feet.
Since the area must be greater than 2f, we get:
\(πf^2 > 2f\)
\(πf > 2\)
\(f > \frac{2}{π}\)

Here, the radius must be greater than \(\frac{2}{π}\) feet.
Case 1 also satisfies Statement 2.
In Case 1, the answer to the question stem is YES.

Case 3: \(r = \frac{2}{3}\) feet = \(\frac{2}{9}\) yard
In this case, area \(= πr^2 = π(\frac{2}{9})^2 = \frac{4}{81}π =\) less than 2/9 square yard
Since the area is LESS than r square yards. the answer to the question stem is NO.

Since the answer is YES in Case 1 but NO in Case 3, INSUFFICIENT.

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